Question

Suppose, we think of fission of a ${}_{26}^{56}Fe$ nucleus into two equal fragments of ${}_{13}^{28}Al$. Is the fission energetically possible? Argue by working out Q of the process. Given m $\left({}_{26}^{56}Fe\right)$ = 55.93494 u and m ${(}_{13}^{28}Al)$ = 27.98191 u.

Answer 1

${}_{26}^{56}Fe\to 2{}_{13}^{28}Al$

$M_1=55.93494u$ is atomic mass of $Fe$.

$M_2=27.98191u$ is atomic mass of $Al$.

$Q$ value is $(M_1-2M_2)\times 931.5MeV$

$(55.93-2\times 27.98)\times 931.5$

$\Rightarrow -26.902MeV$

$Q$ is negative. Hence fission is not possible.